if \(\mathrm{f}(\mathrm{x})=\mathrm{x} \int_{0} \log [(1-\mathrm{t}) /(1+\mathrm{t})] \mathrm{dt}\) then \(\mathrm{f}(1 / 2)-\mathrm{f}[(-1) / 2]\) is equals (a) 0 (b) \((1 / 2)\) (c) \([(-1) / 2]\) (d) 1

To find the closed form of the given integral, we can start by expanding the logarithm using log(a/b) = log(a) - log(b), then integrate the resulting expression with respect to t. The given integral is: \(\int_{0} \log[(1-t)/(1+t)] dt\) Expanding the logarithm, we have: \(\int_{0} (\log(1-t) - \log(1+t)) dt\) Now, we will find the antiderivative of each term separately. For the first term, let u = 1 - t, then du = -dt, and dt = -du. So we get: \(-\int_{0} \log(u) du\) We know that the antiderivative of -log(u) is u * (1 + log(u)). Now, we'll find the antiderivative of the second term: For the second term, let v = 1 + t, then dv = dt, and dt = dv. \(\int_{0} \log(v) dv\) We know that the antiderivative of log(v) is v * (-1 + log(v)). Now, we can write the closed form of the integral: \(\int_{0} \log[(1-t)/(1+t)] dt = u * (1 + log(u)) - v * (-1 + log(v))\) .MapFrom(u,v)->(1-t,1+t) \((1-t)(1+\log(1-t))-(1+t)(-1+\log(1+t))\)

We have found that: \(f(x) = x[(1 - x)(1 + \log(1 - x)) - (1 + x)(-1 + \log(1 + x))]\) Let's evaluate f(x) for x = 1/2: \(f(1/2) = (\frac{1}{2})[((1 - \frac{1}{2})(1 + \log(1 - \frac{1}{2}))-(1 + \frac{1}{2})(- 1 + \log(1 + \frac{1}{2}))]\) \(f(1/2)=\frac{1}{2}(0)\) Now, let's evaluate f(x) for x = -1/2: \(f(-1/2) = (\frac{-1}{2})[((1 - \frac{-1}{2})(1 + \log(1 - \frac{-1}{2}))-(1 + \frac{-1}{2})(- 1 + \log(1 + \frac{-1}{2}))]\) \(f(-1/2)=\frac{-1}{2}(0)\)

Now, we can calculate the difference between f(1/2) and f(-1/2): \(f(1/2) - f(-1/2) = \frac{1}{2}(0) - \frac{-1}{2}(0) = 0\) Comparing this result with the given options, we can see that: \(f(1/2) - f(-1/2) = 0\) This matches option (a). Therefore, the correct answer is: (a) 0